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Lyrx [107]
3 years ago
6

A reaction between liquid reactants takes place at in a sealed, evacuated vessel with a measured volume of . Measurements show t

hat the reaction produced of dinitrogen monoxide gas. Calculate the pressure of dinitrogen monoxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to significant digits.
Chemistry
1 answer:
Oxana [17]3 years ago
6 0

This question is incomplete, the complete question;

A reaction between liquid reactants takes place at 16°C in a sealed, evacuated vessel with a measured volume of 10.0 L . Measurements show that the reaction produced 37.g of dinitrogen monoxide gas.

Calculate the pressure of dinitrogen monoxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.

Answer:

Explanation:

Given that;

Volume = 10.0 L

Temperature T = 16°C = ( 16 + 273) = 289 K

we know that  molar mass of N₂O = 44 g/mol

number number of moles n will be: weight/molar mass = 37/44 = 0.8409 moles

Universal Gas constant R = 0.0821 atm.mol⁻¹.k⁻¹

pressure p = ?

from the ideal gas law

PV = nRT

P = nRT / n

so we substitute

P = (0.8409 mole × 0.0821 atm.mol⁻¹.k⁻¹ × 289 K) / 10.0 l

P = 19.95195 / 10.0

P = 1.9952 atm

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                              ----------------------------------
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The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

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